使序列有序的最少交换次数（minimum swaps）

题目描述：（minimum swaps）

Given a sequence, we have to find the minimum no of swaps required to sort the sequence.

 

分析：

formula：  no. of elements out of place  -  "cycles" in the sequence

 

A cycle is a set of elements, each of which is in the place of another.  So in example sequences { 2, 1, 4, 3}, there are two cycles: {1, 2} and {3, 4}.  1 is in the place where 2 needs to go, and 2 is in the place where 1 needs to go 1, so they are a cycle; likewise with 3 and 4.

 

The sequences {3, 2, 1, 5, 6, 8, 4, 7 }also has two cycles: 3 is in the place of 1 which is in the place of 3, so {1, 3} is a cycle; 2 is in its proper place; and 4 is in the place of 7 which is in the place of 8 in place of 6 in place of 5 in place of 4, so {4, 7, 8, 6, 5} is a cycle.  There are seven elements out of place in two cycles, so five swaps are needed.

 

实现：

e.g. { 2, 3, 1, 5, 6, 4}

231564 -> 6 mismatch
two cycles -> 123 and 456
swap 1,2 then 2,3, then 4,5 then 5,6 -> 4 swaps to sort
 
Probably the easiest algorithm would be to use a bitarray. Initialize it to 0, then start at the first 0. Swap the number there to the right place and put a 1 there. Continue until the current place holds the right number. Then move on to the next 0
 
Example:
231564
000000
-> swap 2,3
321564
010000
-> swap 3,1
123564
111000
-> continue at next 0; swap 5,6
123654
111010
-> swap 6,4
123456
111111
-> bitarray is all 1's, so we're done.

 

#include <iostream>
#include <algorithm>

using namespace std;


template <typename T>
int GetMinimumSwapsForSorted(T seq[], int n)
{
	bool* right_place_flag = new bool[n];
	T* sorted_seq = new T[n];
	int p ,q;
	////
	copy(seq, seq + n, sorted_seq);
	sort(sorted_seq, sorted_seq + n);    ////可采用效率更高的排序算法
	// 初始化bitarray
	for(int i = 0; i < n; i++)
	{
		if(seq[i] != sorted_seq[i])
			right_place_flag[i] = false;
		else
			right_place_flag[i] = true;
	}
	////
	p = 0;
	int minimumswap = 0;
	while(1)
	{
		while(right_place_flag[p])
			p++;
		q = p + 1;
		// 在已排好序的数组中找到和未排好序的元素，此种找法只对无重复序列能得出minimum swaps
		while(q < n)
		{
			if(!right_place_flag[q] && sorted_seq[q] == seq[p])
				break;
			q++;
		}

		if(q >= n || p >= n)
			break;
		right_place_flag[q] = true;
		// 对调后正好在排好序数组的位置上，则bitarray中的那位也置为true
		if(seq[q] == sorted_seq[p])
			right_place_flag[p] = true;
		swap(seq[p], seq[q]);
		minimumswap++;
	}

	delete[] sorted_seq;
	delete[] right_place_flag;

	return minimumswap;
}


int _tmain(int argc, _TCHAR* argv[])
{
	int seq[] = {3, 2, 1, 5, 6, 8, 4, 7 };//{2,3,1,5,6,4};//{2,3,2,4,7,6,3,5};
	int n = sizeof(seq) / sizeof(int);
	cout<<"minimum swaps : "<<GetMinimumSwapsForSorted(seq, n)<<endl;

	system("pause");
	return 0;
}